Vrftsjtryk

I'm looking for a smooth map from the 3-torus to $SU(2)$, i.e. some $f:mathbb{T}^{3}to SU(2)$, that is injective near identity, $f^{-1}(1)={p}$. I know that the homotopy classes of maps from $mathbb{T}^{3}to S^{3}simeq SU(2)$ are labeled by $mathbb{Z}$, but I need to find an explicit example (I think of the $n=1$ class) for my work.

If there's an easy, explicit construction, I'm all for it. Below I've tabulated my attempt at first embedding $mathbb{T}^{2}$ in $mathbb{R}^{3}$ and projecting to the unit sphere, hoping that the same idea works with $mathbb{T}^{3}$ in $mathbb{R}^{4}$.

Because I can't think in four dimensions, my initial idea was to knock off one dimension from both spaces and try to find a map from a 2-torus to $S^{2}$ that satisfies this.

So I've got a Torus parameterized as:
$$ x=(R+rcostheta)cosphi-R,hspace{.5cm} y=(R+rcostheta)sinphi, hspace{.5cm}z=rsintheta$$
Crucially, this torus has the origin inside of it, so that when I project to the unit sphere
$$ (x, y, z)mapsto frac{(x, y, z)}{sqrt{x^{2}+y^{2}+z^{2}}}$$
only one point ($theta=0, phi=0$) is mapped to $(1, 0, 0)$ on the unit sphere. So I can map $mathbb{T}^{2}to S^{2}$ by:
$$(theta, phi) mapsto frac{(x, y, z)}{sqrt{x^{2}+y^{2}+z^{2}}}$$

(1) I'm not sure that this is even a smooth map, even though it seems like it should be. The issue is that one ends up with a $sqrt{...+costheta}$ in the denominator, which contains arbitrarily high powers of $costheta$. (Ultimately I'm using this for a physics application which needs a local formulation, and $theta$ will be a momentum, so I need there to be a finite maximum power of $costheta$, $sintheta$, $cosphi$, etc)

(2) If this lower dimensional case is smooth, can the same idea be extended to a map $mathbb{T}^{3}to SU(2)$? Can one parameterize $mathbb{T}^{3}$ in a similar way to $mathbb{T}^{2}$, then project to the unit sphere, and be done with it?

(3) Is there an altogether better way to construct a map from $mathbb{T}^{3}to SU(2)$ that avoids all of this messiness?

First of all, $SU(2)$ is diffeomorphic to $S^3$ (I can write down an explicit diffeomorphism if you like), so I will be constructing a map $T^3to S^3$. I will identify $S^3$ with the 1-point compactification of $R^3$, $S^3=R^3cup {infty}$; this is done via a stereographic projection.

The idea is that $T^3$ is homeomorphic to the cube $Q=[-1,1]^3$ in 3-space with opposite faces identified via translations. Thus, I will construct a continuous map $f: Qto S^3$ which is constant (equal to $infty$) on the boundary and is a homeomorphism (to $R^3$) on the interior of $Q$.

The map $f$, of course, respects the identification of the opposite faces on the boundary of $Q$, hence, defines the required map $T^3to S^3$.

Let $S$ denote the unit sphere in $R^4$ centered at $e_4=(0,0,0,1)$, it is given by the equation
$$
x_1^2 +x_2^2 + x_3^2 + (1-x_4)^2=1.
$$

I will construct $f$ as the composition of three explicit maps.

1. An explicit homeomorphism $h_1$ from $Q$ to the unit ball $B$ centered at the origin, see for instance here.




2. The homeomorphism $h_2$ from $B$ to the lower hemisphere $H$ in $S$:
   $$
   h_2(x_1,x_2,x_3)= left(x_1,x_2,x_3, 1- sqrt{1- x_1^2 -x_2^2 - x_3^2}right).
   $$
   This map sends the unit sphere
   $$
   x_1^2 +x_2^2 + x_3^2=1
   $$
   to the boundary of the hemisphere $H$, i.e. the intersection of the sphere $S$ with the hyperplane $x_4=1$.




3. The stereographic projection $h_3$ with the center at $e_4$ and defined on the slab
   $$
   P= {(x_1,x_2,x_3,x_4): 0le x_4le 1}.
   $$
   The map $h_3$ sends every point $p=(x_1,x_2,x_3,x_4)in P$ satisfying
   $$
   0le x_4< 1
   $$
   to the intersection point of the line through $p$ and $e_4$ with the hyperplane $x_4=0$, and sends points with $x_4=1$ to
   $infty$. In particular, $h_3$ sends the boundary of the hemisphere $H$ to $infty$. It sends $H$ minus its boundary sphere, homeomorphically to $R^3$

Now, $f= h_3circ h_2 circ h_1$.